Sunday, January 30, 2022

Even more about normal subgroups and examples of factor groups

Normal subgroups are an important part of the structure of a group. We already have the definition.

 

If H is a subgroup contained in G and for every element x in G, xH = Hx, then H is normal.

 

We have also defined G/H as a new group called the factor group.

 

Every group G that has more than one element must have at least two subgroups, G itself and the identity element e by itself. Both of these subgroups are normal. G/G is a one element group, which means (e,*) and G/e = G.


Lemma: If G has order 2n and H has order n, H must be normal and G/H is isomorphic to Z2.


Proof: The elements of G that are not in H are denoted as G-H. If h is in H, hH = Hh = H, because the subgroup is closed under the group operation. If x is in G-H, xH = Hx = G-H.


Every group of order 2 is isomorphic to Z2 because if we make the Cayley table, it is a 2x2 matrix with two elements that follows the rule of Latin square, that every row and every column has exactly one copy of each element.


+ | 0 | 1 

0 | 0 | 1

1 | 1 | 0


We have seen that (nZ, +) is a subgroup of Z. Z/nZ = Zn. This factor group has n infinite cosets we can call nZ, nZ+1, nZ+2,... up to nZ+(n-1). We take an element from each coset and make it the representative. Usually the representatives are the integers from 0 to n-1, but on a clock, for example, we use 12 instead of 0 in Z12.

 

What does (R/Z, +) look like? You can think of it as the half closed, half open interval [0, 1), or you could visualize the fact that 1 is mapped onto 0, so it is a closed loop. 

 

What does (Q/Z, +) look like? It looks like (R/Z, +) to the naked eye, and no amount of magnification will change that, but there are an infinity of numbers missing from the interval.

 

How about (R/Q, +)? This one is trippy. All we would see is the single point 0, but the irrationals would all be sent to an infinite number of infinitesimal values. If you want to learn more about the topic, follow this link to the Wikipedia page.   

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